Question

A ballerina begins a tour jet (Figure a) with angular speed ${\mathbf{\omega }}_{\mathbf{i}}$and a rotational inertia consisting of two parts : role="math" localid="1661005078220" ${\mathbf{I}}_{\mathbf{leg}}$= 1.44 kg.m² for her leg extended outward at angle $\mathbf{\theta }$= 90.0$\mathbf{°}$to her body and ${\mathbf{I}}_{\mathbf{trunk}}$= 0.660 kg.m² for the rest of her body (primarily her trunk). Near her maximum height she holds both legs at angle$\mathbf{30}\mathbf{.}\mathbf{0}\mathbf{°}$to her body and has angular speed${\mathbf{\omega }}_{\mathbf{f}}$(Figure b). Assuming that Ihas not changed, what is the ratio$\raisebox{1ex}{${\mathbf{\omega }}_{\mathbf{f}}$}\!\left/ \!\raisebox{-1ex}{${\mathbf{\omega }}_{\mathbf{i}}$}\right.$ ?

(a) Initial phase of a tour jet: large rotational inertia and small angular speed. (b) Later phase: smaller rotational inertia and larger angular speed.

Short answer

Ratio of final to initial angular velocities is $\frac{{\mathrm{\omega }}_{\mathrm{f}}}{{\mathrm{\omega }}_{\mathrm{i}}}=1.52$.

Step-by-step solution

Step 1: Given

$\begin{array}{l}{\mathrm{I}}_{\mathrm{leg}}=1.44\text{kg}.{\text{m}}^2\\ {\mathrm{I}}_{\mathrm{trunk}}=0.660\text{kg}.{\text{m}}^2\end{array}$

Determining the concept

Calculate initial rotational inertia when leg extending at ${\mathbf{\text{90}}}^{\mathbf{\text{0}}}$and final rotational inertia when leg extending at${\mathbf{\text{30}}}^{\mathbf{\text{0}}}$. Then, apply law of conservation of momentum to find the ratio of final to initial angular velocities. According tothe conservation of momentum, momentum of a system is constant if no external forces are acting on the system.

Formula are as follow:

1. Initial angular momentum = Final angular momentum
2. role="math" localid="1661005530030" ${\mathbf{I}}_{\mathbf{i}}\mathbf{=}{\mathbf{I}}_{\mathbf{trunk}}\mathbf{+}{\mathbf{I}}_{\mathbf{leg}}$

Where,${\mathbf{I}}_{\mathbf{trunk}}$is moment of inertia of trunk and${\mathbf{I}}_{\mathbf{leg}}$is moment of inertia of leg

Determining the ratio of final to initial angular velocities

Initial rotational inertia when both leg extending ${90}^0$outward,

$\begin{array}{l}{\mathrm{I}}_{\mathrm{i}}={\mathrm{I}}_{\mathrm{trunk}}+{\mathrm{I}}_{\mathrm{leg}}\\ {\mathrm{I}}_{\mathrm{i}}=0.660+1.44=2.10{\text{kg.m}}^{\text{2}}\end{array}$

Final rotational inertia when both leg extending${30}^0$ outward,

$\begin{array}{l}{\mathrm{I}}_{\mathrm{f}}={\mathrm{I}}_{\mathrm{trunk}}+{\mathrm{I}}_{\mathrm{eff}}\\ {\mathrm{I}}_{\mathrm{f}}={\mathrm{I}}_{\mathrm{trunk}}+2{\mathrm{I}}_{\mathrm{leg}}{\sin }^2\left(\mathrm{\theta }\right)\end{array}$

The factor $2{\sin }^2\mathrm{\theta }$arises from the fact that there are two legs and each one of them is stretched at an angle $30°$. So, effective length from the axis of rotation would be,

$\begin{array}{l}{\mathrm{l}}_{\mathrm{eff}}=\mathrm{L}\times \mathrm{sin\theta }\\ {\mathrm{I}}_{\mathrm{eff}}={\mathrm{ml}}_{\mathrm{eff}}^2={\mathrm{mL}}^2{\sin }^2\left(\mathrm{\theta }\right)\\ {\mathrm{I}}_{\mathrm{eff}}=2{\mathrm{mL}}^2{\sin }^2\left(\mathrm{\theta }\right)\\ {\mathrm{I}}_{\mathrm{eff}}=2{\mathrm{I}}_{\mathrm{leg}}{\sin }^2\left(\mathrm{\theta }\right)\end{array}$

${\mathrm{I}}_{\mathrm{f}}=0.660+2\times 1.44{\sin }^2\left(30\right)=1.38{\text{kg.m}}^{\text{2}}$

Now, according to law of conservation of momentum,

$\begin{array}{c}{\mathrm{L}}_{\mathrm{i}}={\mathrm{L}}_{\mathrm{f}}\\ {\mathrm{I}}_{\mathrm{i}}{\mathrm{\omega }}_{\mathrm{i}}={\mathrm{I}}_{\mathrm{f}}{\mathrm{\omega }}_{\mathrm{f}}\end{array}$

Hence,

$\frac{{\mathrm{\omega }}_{\mathrm{f}}}{{\mathrm{\omega }}_{\mathrm{i}}}=\frac{{\mathrm{I}}_{\mathrm{i}}}{{\mathrm{I}}_{\mathrm{f}}}=\frac{2.10}{1.38}=1.52$

Hence, ratio of final to initial angular velocities is $\frac{{\mathrm{\omega }}_{\mathrm{f}}}{{\mathrm{\omega }}_{\mathrm{i}}}=1.52$.

Therefore, the total rotational inertia of the system can be found. Using the law of conservation of momentum, ratio of angular velocities can be found.